Finding the value of $\lim_{a\to \infty}\int_0^1 a^x x^a \,dx$

Question

I'm trying to find the value of $$\lim_{a\to \infty}\int_0^1 a^x x^a \,dx$$ My attempt: Let $\epsilon >0$ be given.

$ x\mapsto a^{x}$ is continuous at $ 1$ so there is a $d_a\in ( 0,1)$ such that $|a^{x} -a|< \epsilon $ for all $ x\in [d_a,1]$. WLOG, let $d_a<1/2$.

$ |\int _{0}^{1} x^{a} a^{x} \ dx-\ \int _{0}^{1} \ ax^{a} |=|\int _{0}^{1}\left( a^{x} -a\right) x^{a} \ dx|\leq |\int _{0}^{d}\left( a^{x} -a\right) x^{a} \ dx|+|\int _{d}^{1}\left( a^{x} -a\right) x^{a} \ dx|$

\begin{align*} \left|\int _{0}^{1} x^{a} a^{x} \ dx-\ \int _{0}^{1} \ ax^{a} \right| & \leq \left|\int _{0}^{d_a}\left( a^{x} -a\right) x^{a} \ dx\right|+\left|\int _{d_a}^{1}\left( a^{x} -a\right) x^{a} \ dx\right|\\ & \leq \int _{0}^{d_a}\left( a -a^{x}\right) x^{a} \ dx+\epsilon \left|\int _{d_a}^{1} x^{a} \ dx\right|\\ & \leq \int _{0}^{d_a}\left( a -a^{x}\right) x^{a} \ dx+\epsilon \\ & \leq \int _{0}^{1/2} a(1/2)^{a} \ dx-a\int _{0}^{d_a} x^{a} dx+\epsilon \\ & \leq a(1/2)^{a} +\epsilon \end{align*} $0\leq \lim _{a\rightarrow \infty }\inf |\int _{0}^{1} x^{a} a^{x} \ dx-\ \int _{0}^{1} \ ax^{a} |\leq \lim _{a\rightarrow \infty }\sup |\int _{0}^{1} x^{a} a^{x} \ dx-\ \int _{0}^{1} \ ax^{a} |\leq \epsilon $

Since this is true for every $\epsilon >0,$it follows that $ \lim _{a\rightarrow \infty }\int _{0}^{1} x^{a} a^{x} \ dx-\ \int _{0}^{1} \ ax^{a} =0$.

Is my proof correct? Thanks.

Answer 1

An Upper Bound

For $a\ge1$, $$ \begin{align} \int_0^1a^xx^a\,\mathrm{d}x &\le\int_0^1ax^a\,\mathrm{d}x\tag{1a}\\ &=\frac{a}{a+1}\tag{1b} \end{align} $$ Explanation:
$\text{(1a)}$: $a^x\le a$
$\text{(1b)}$: evaluate the integral

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A Lower Bound $$ \begin{align} \int_0^1a^xx^a\,\mathrm{d}x &=\frac{a}{a+1}\int_0^1a^{x-1}\,(a+1)x^a\,\mathrm{d}x\tag{2a}\\ &\ge\frac{a}{a+1}a^{\int_0^1(x-1)\,(a+1)x^a\,\mathrm{d}x}\tag{2b}\\ &=\frac{a}{a+1}a^{-\frac1{a+2}}\tag{2c} \end{align} $$ Explanation:
$\text{(2a):}$ factor $\frac{a}{a+1}$ out of the integral
$\text{(2b):}$ Jensen's Inequality
$\phantom{\text{(2b):}}$ $a^x$ is convex and $\int_0^1(a+1)x^a\,\mathrm{d}x=1$
$\text{(2c):}$ evaluate the integral

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Squeeze The Limit

For $a\ge1$, $(1)$ and $(2)$ give $$ \frac{a}{a+1}a^{-\frac1{a+2}}\le\int_0^1a^xx^a\,\mathrm{d}x\le\frac{a}{a+1}\tag3 $$ and the Squeeze Theorem gives $$ \lim_{a\to\infty}\int_0^1a^xx^a\,\mathrm{d}x=1\tag4 $$ Here is a graph plotting the bounds given in $(3)$ and also the lower bound, in red, given in Paramanand Singh's nice answer (both our upper bounds are the same):

Answer 2

Here is a slightly different solution base on the change of variable $u=1-x$.

First the easy bound $$I_a=\int^1_0e^{x\log a}x^a\,dx\leq a\int^1_0x^a\,dx=\frac{a}{1+a}\xrightarrow{a\rightarrow\infty}1$$

The change of variables $u=1-x$ yields $$I_a=\int^1_0a^{1-u}(1-u)^a\,du$$

Since $e^{\tfrac{u}{1-u}}\geq 1+\frac{u}{1-u}=\frac{1}{1-u}$, we have that $(1-u)^a\geq e^{-\tfrac{au}{1-u}}$ for $0\leq u \leq 1$ (both RHS and LHS take value $0$ when $u\rightarrow1-$). Hence, for all $a>e$ $$\begin{align} I_a&=\int^1_0a^{1-u}(1-u)^a\,du\geq a\int^1_0a^{-u}\exp\big(\tfrac{-au}{1-u}\big)\,du\\ &\geq a\int^{1/\log a}_0e^{-u\big(\log a+\tfrac{a}{1-u}\big)}\,du\\ &\geq a\int^{1/\log a}_0e^{-u\big(\log a+\tfrac{a}{1-1/\log a}\big)}\,du\\ &=\frac{1}{\frac{\log a}{a}+\frac{1}{1-1/\log a}}\Big(1-e^{-\tfrac{1}{\log a}\big(\log a+\tfrac{a}{1-1/\log a}\big)}\Big)\xrightarrow{a\rightarrow\infty}1 \end{align} $$

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Comment: the bound $(1-u)^a\geq e^{-\tfrac{au}{1-u}}$ seems to be optimal to get the right limit. Other bounds, for example by using Bernoulli's inequality $(1-u)^a\geq 1-au\geq0$ for $0\leq u\leq 1/a$, fall short: $$I_a\geq \int^{1/a}_0a^{1-u}(1-au)\,du\xrightarrow{a\rightarrow\infty}\frac12$$

Answer 3

Let's observe that $$\int_0^1(a-a^x)x^a\,dx$$ is slightly easier to handle. The integrand equals $$x^a(1-x)a^y\log a$$ for some $y\in(x,1)$ via mean value theorem and hence does not exceed $$x^a(1-x)a\log a\tag{1}$$ We can also use the fundamental inequality $e^x\geq 1+x$ for all real $x$ to get $$a-a^x=a\left(1-e^{(x-1)\log a}\right)\leq a(1-x)\log a$$ to reach the bound $(1)$ for the integrand (this is based on suggestion from robjohn in comments).

Upon integrating the expression $(1)$ on $[0,1]$ we get $$\frac{a\log a}{(a+1)(a+2)}$$ and it tends to $0$. It follows that integral at beginning of this answer tends to $0$ and the desired limit is thus $$\lim_{a\to\infty}a\int_0^1 x^a\,dx=1$$

Answer 4

Here is another method: Substituting $y = x^a$, or equivalent, $x = y^{1/a}$, the integral boils down to

\begin{align*} I_a := \int_{0}^{1} a^x x^a \, \mathrm{d}x &= \int_{0}^{1} a^{y^{1/a} - 1} y^{1/a} \, \mathrm{d}y \\ &= \int_{0}^{1} y^{1/a} \exp\bigl( (y^{1/a} - 1)\log a \bigr) \, \mathrm{d}y. \end{align*}

Since the integrand is bounded between $0$ and $1$ (provided $a \geq 1$), by the dominated convergence theorem, this converges to:

\begin{align*} \lim_{a\to\infty} I_a &= \int_{0}^{1} \lim_{a \to \infty} y^{1/a} \exp\bigl( (y^{1/a} - 1)\log a \bigr) \, \mathrm{d}y \\ &= \int_{0}^{1} \, \mathrm{d}y = 1. \end{align*}

Answer 5

There has been a bit of activity on this question recently, so I considered an approach significantly different from my previous answer; different enough to warrant another answer rather than amending the existing answer.

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Preliminary Limit

Let $u\in(0,1)$ and $a\ge1$. Then $$ u^{\frac1{a+1}}\le1\tag1 $$ Furthermore, Bernoulli's Inequality gives $$ u^{\frac1{a+1}}=(1+(1/u-1))^{-\frac1{a+1}}\ge1-\frac{1/u-1}{a+1}\tag2 $$ Thus, $$ -\frac{1/u-1}{a+1}\le u^{\frac1{a+1}}-1\le0\tag3 $$ and since $a\ge1$, $$ a^{-\frac{1/u-1}{a+1}}\le a^{u^{\frac1{a+1}}-1}\le1\tag4 $$ Therefore, by the fact that $\lim\limits_{a\to\infty}a^{\frac1{a+1}}=1$ and the Squeeze Theorem, we have $$ \lim_{a\to\infty}a^{u^{\frac1{a+1}}-1}=1\tag5 $$

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The Main Limit $$ \begin{align} \lim_{a\to \infty}\int_0^1a^xx^a\,\mathrm{d}x &=\lim_{a\to \infty}\frac1{a+1}\int_0^1a^{u^{\frac1{a+1}}}\,\mathrm{d}u\tag{6a}\\ &=\lim_{a\to \infty}\frac{a}{a+1}\int_0^1a^{u^\frac1{a+1}-1}\,\mathrm{d}u\tag{6b}\\ &=\lim_{a\to \infty}\frac{a}{a+1}\ \lim_{a\to \infty}\int_0^1a^{u^\frac1{a+1}-1}\,\mathrm{d}u\tag{6c}\\ &=1\cdot\int_0^11\,\mathrm{d}u\tag{6d}\\[6pt] &=1\tag{6e} \end{align} $$ Explanation:
$\text{(6a):}$ substitute $u=x^{a+1}$, then $x=u^{\frac1{a+1}}$ and $x^a\,\mathrm{d}x=\frac1{a+1}\,\mathrm{d}u$
$\text{(6b):}$ bring a factor of $a$ outside the integral
$\text{(6c):}$ the limit of a product is the product of the limits
$\text{(6d):}$ apply $(4)$ and $(5)$ and Dominated Convergence
$\text{(6e):}$ evaluate the integral

Answer 6

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ As $\ds{a \to \infty}$, the $\ds{a^{x}x^{a}}$ main contribution to the integral happens to be around $\ds{x = 1}$. Therefore, \begin{align} &\bbox[5px,#ddf]{% \lim_{a \to \infty}\int_{0}^{1}a^{x}x^{a}\,\dd x} = \lim_{a \to \infty}\int_{0}^{1}a^{1 - x}\,\, \pars{1 - x}^{a}\,\dd x \\[5mm] = & \ \lim_{a \to \infty}\int_{0}^{1} \exp\pars{\pars{1 - x}\ln\pars{a} + a\ln\pars{1 - x}}\,\dd x \\[2mm] = & \ \lim_{a \to \infty}\,\,\,\overbrace{\int_{0}^{\infty} \exp\pars{\ln\pars{a} - \bracks{a + \ln\pars{a}}x} \,\dd x}^{\ds{Laplace's\ Method}} \ \\[5mm] = & \lim_{a \to \infty}\ {a \over a + \ln\pars{a}} = \bbx{\large 1} \end{align}